FEUERBACH POINT PROBLEM

Lemma 1

Let ABC be a triangle. The Oi line intersects AB at C'.The circle (Oa) with diameter AI intersects the circumcircle again at A". The line C'A" intersects (Oa) again at Ca.Ca is the orthogonal projection of A on IC.

Lemma 2

Let ABC be a triangle and Ba, Ca the orthogonal projeections of A on IB, IC, resp. The line BaCa intersects AC, AB at B'a, C'a, resp. B'a,C'a are the midpoints of AC, AB, resp.

Lemma 3

Let ABC be a triangle. Denote C' = the midpoint of AB, Ma, Mb = the midpoints of AI, BI, resp. and Ba = the orthogonal projection of A on BI. The points Ma, C', Mb, Ba are concyclic. The circle is the NPC of AIB.

Problem
By David Nguyen

Rewritten with cyclic notation
Let ABC be a triangle. The OI line intersects BC, CA, AB at A', B', C', resp.  
Denote
(Oa) = the circle with diameter AI
A" = the 2nd intersection of the circumcircle and (Oa)
Ba = the 2nd intersection of B'A" and (Oa)
Ca = the 2nd intersection of C'A" and (Oa)
B'a = BaCa ∩ AC  , C'a = BaCa ∩ AB

(Oab) = the circumcircle of OaB'aCa
(Oac)= the circumcircle of OaBaC'a
The 2nd intersection of (Oab), (Oac) is the Feuerbach point

Proof
By the lemmata, the problem is equivalent to the following problem
Let ABC be a triangle and A'B'C' the medial triangle
Denote
A", B", C" = the midpoints of IA, IB, IC, resp.
The circumcircles of A'B"C", B'C"A", C'A"B" concur at the Feuerbach point
True, since the circumcircles of A'B"C", B'C"A", C'A"B" are the NPCs of IBC ICA, IAB resp. and they concur at the Poncelet point of ABCI = the Feuerbach point.

Mail Antreas P. Hatzipolakis

PANAKIS' PSEUDOISOSCELES TRIANGLE

Let ABC be a trangle and D1, D2, D3 the feet of the internal angle bisectors [D1D2D3 = the cevian triangle of the incenter I]
Prove that the triangle D1D2D3 can be isosceles without ABC being isosceles.
Ioannis Panakis, Plane Trigonometry, vol. B, Athens (1973), p. 110 [in Greek]

I call this triangle ABC as Panakis pseudoisosceles triangle

Properties of ABC (in the same book pp. 109-111)
1. The A, D1, D2, D3 are concyclic [the cevian circle of I passes through A]
2. (a + b + c)*(-a^2 + b^2 + c^2) + abc = 0
3. a / (b + c) = b / (c + a) + c / (a + b)
4. (r1 - r) / (r1 + r) = ((r2 - r) / (r2 + r)) + ((r3 - r) / (r3 + r))
where r = the inradius and r1, r2, r3 the exradii.

PDF of the pages of the book Panakis

Mail Antreas P. Hatzipolakis

THEBAULT'S PSEUDOISOSCELES TRIANGLE

Victor Thebault published the following theorem as an exercise:

Si le cercle qui passe par les pieds des bissectrices intérieures d'un triangle est tangent à l'un des côtés, le triangle est isocèle, et réciproquement. 
(If the circle passing through the feet of the interior bisectors of a triangle is tangent to one of the sides, the triangle is isosceles, and vice versa.)

Solution by (A.M.)  [false]
Journal de mathématiques élémentaires.  
75e Annee - No 1 -  1er Octobre 1950, p. 3, #14250

Joseph Andersonn proved that the triangle is not necessarily isosceles.
CERCLE PASSANT PAR LES PIEDS DES BISSECTRICES INTÉRIEURES D'UN TRIANGLE ET TANGENT À L'UN DES CÔTÉS
par A. Monjallon. 
Journal de mathématiques élémentaires.  
75e Annee - No 20, 15 Juillet 1951, pp. 153 - 4

CERCLE PASSANT PAR LES PIEDS DES BISSECTRICES INTÉRIEURES D'UN TRIANGLE ET TANGENT À L'UN DES CÔTÉS
par Rene Blanchard. 
Journal de mathématiques élémentaires.  
76e Annee - No 4, 15 Novembre 1951, pp. 25 - 6 

PDF File Victor Thebault

Francisco Javier García Capitán's Solution (in Spanish)
wrong-thebault

Mail Antreas P. Hatzipolakis